Problem: You have found the following ages (in years) of 5 zebras. Those zebras were randomly selected from the 37 zebras at your local zoo: $ 13,\enspace 1,\enspace 10,\enspace 37,\enspace 7$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 37 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{13 + 1 + 10 + 37 + 7}{{5}} = {13.6\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {0.36} + {158.76} + {12.96} + {547.56} + {43.56}} {{5 - 1}} $ {s^2} = \dfrac{{763.2}}{{4}} = {190.8\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{190.8\text{ years}^2}} = {13.8\text{ years}} $ We can estimate that the average zebra at the zoo is 13.6 years old. There is also a standard deviation of 13.8 years.